To find the equation of the second street, we need to ensure it is parallel to the first street and passes through the given point [tex]\((-2, 4)\)[/tex].
1. Determine the slope of the first street:
The equation of the first street is [tex]\(2x + y = 2\)[/tex]. We need to put this equation in slope-intercept form, [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] is the slope and [tex]\(b\)[/tex] is the y-intercept.
Solving for [tex]\(y\)[/tex]:
[tex]\[
2x + y = 2 \implies y = -2x + 2
\][/tex]
The slope of the first street is [tex]\(m = -2\)[/tex].
2. Ensure the second street is parallel:
For the second street to be parallel to the first street, it must have the same slope, [tex]\(m = -2\)[/tex].
3. Use the point-slope form with the slope and given point:
The point-slope form of a line is given by:
[tex]\[
y - y_1 = m(x - x_1)
\][/tex]
Here, [tex]\(m = -2\)[/tex], [tex]\(x_1 = -2\)[/tex], and [tex]\(y_1 = 4\)[/tex]. Plugging these into the form gives:
[tex]\[
y - 4 = -2(x + 2)
\][/tex]
4. Simplify the equation:
Distribute and simplify:
[tex]\[
y - 4 = -2x - 4 \implies y = -2x
\][/tex]
5. Convert to standard form:
Standard form of a linear equation is [tex]\(Ax + By = C\)[/tex]. Rearrange [tex]\(y = -2x + 8\)[/tex] to:
[tex]\[
-2x + y = 8 \implies 2x - y = -8 \quad (\text{Multiply entire equation by } -1)
\][/tex]
After these steps, we arrive at the equation in standard form. Among the given choices, the correct one is:
[tex]\[ x + y = 2 \][/tex]
Therefore, the equation of the second street is:
[tex]\[ \boxed{x + y = 2} \][/tex]
